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Linear Algebra




                    Notes                                        p = (x – c ) ... (x – c )
                                                                        1      k
                                   where c ,..., c  are distinct elements of F.
                                         1   k
                                   Proof: We have noted earlier that, if T is diagonalizable, its minimal polynomial is a product of
                                   distinct linear factors. To prove the converse, let  W be  the subspace spanned by  all of the
                                   characteristic vectors of T, and suppose W  V. By the lemma used in the proof of Theorem 1,
                                   there is a vector  not in W and a characteristic value c  of T such that the vector
                                                                              j
                                                                     = (T – c I)
                                                                           j
                                   lies in W. Since  is in W,
                                                                    =   + ... + 
                                                                       1      k
                                   where T  = c  , 1  i  k, and therefore the vector
                                          i  i  i
                                                              h(T) = h(c )  + ... + h(c )
                                                                       1  1      k  k
                                   is in W, for every polynomial h.
                                   Now p = (x – c )q, for some polynomial q. Also
                                              j
                                                                  q – q(c ) = (x – c )h
                                                                       j      j
                                   We have
                                                          q(T) – q(c ) = h(T)(T – c I) = h(T’)
                                                                  j           j
                                   But h(T) is in W and, since

                                                               0 = p(T) = (T – c I)q(T)
                                                                             j
                                   the vector q(T) is in W. Therefore, q(c ) is in W. Since  is not in W, we have q(c ) = 0. That
                                                                  j                                   j
                                   contradicts the fact that p has distinct roots.
                                   In addition to being an elegant result, Theorem 2 is useful in a computational way. Suppose we
                                   have a linear operator T, represented by the matrix A in some ordered basis, and we wish to
                                   know if T is diagonalizable. We compute the characteristic polynomial f. If we can factor f:
                                                                        1 d
                                                                f   (x c  )  (x c  )  k d
                                                                    
                                                                             
                                                                       1        k
                                   we have two different methods for determining whether or not T is diagonalizable. One method
                                   is to see whether (for each i) we can find d  independent characteristic vectors associated with the
                                                                    i
                                   characteristic value c . The other method is to check whether or not (T – c I)  (T – c I) is the zero
                                                   i                                        1        k
                                   operator.
                                   Theorem 1 provides a different proof of the Cayley-Hamilton theorem. That theorem is easy for
                                   a  triangular  matrix.  Hence,  via  Theorem  1,  we  obtain  the result  for  any  matrix over  an
                                   algebraically closed field. Any field is a subfield of an algebraically closed field. If one knows
                                   that result, one obtains a proof of the Cayley-Hamilton theorem for matrices over any field. If
                                   we at least admit into our discussion the Fundamental Theorem of Algebra (the complex number
                                   field is algebraically closed), then Theorem 1 provides a proof of the Cayley-Hamilton theorem
                                   for complex matrices, and that proof is independent of the one which we gave earlier.

                                   Self Assessment

                                                                  2
                                   1.  Let T be the linear operator on R , the matrix of which in the standard ordered basis is
                                                                            1
                                                                        1  
                                                                    A      
                                                                        2  2  



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